A steady current $I$ goes through a wire loop $P Q R$ having shape of a right angle triangle with $P Q=3 x, P R=4 x$ and $Q R=5 x$. If the magnitude of the magnetic field at $P$ due to this loop is $k\left(\frac{\mu_0 I}{48 \pi x}\right)$, find the value of $k$.

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Aicharya · Accepted Answer

Magnetic field at point $P$ due to wires $R P$ and $R Q$ is zero. Only wire $Q R$ will produce magnetic field at $P$. $r=3 x \cos 37^{\circ}=(3 x)\left(\frac{4}{5}\right)=\frac{12 x}{5}$

Using the concept of area of triangle $\begin{aligned} & \frac{1}{2} \times P D \times 5 x=\frac{1}{2} \times 3 x \times 4 x \\ & \therefore P D=\frac{12 x}{5} \end{aligned}$

$\begin{aligned} & Q D=\sqrt{(P Q)^2-(P D)^2}=\sqrt{9 x^2-\frac{144 x^2}{25}}=\frac{9 x}{5} \\ & \text { and } D R=5 x-\frac{9 x}{5}=\frac{16 x}{5} \end{aligned}$ Magnetic field at $P$ due to current elements $P Q$ and $P R$ is zero as the point $P$ is on the conductor. Therefore, magnetic field at $\mathrm{P}$ due to current element $\mathrm{QR}$ is $\begin{aligned} B & =\frac{\mu_0 I}{4 \pi P D}\left(\sin \phi_1+\sin \phi_2\right) \\ B & =\frac{\mu_0 I \times 5}{4 \pi \times 12 x}\left(\frac{9 x / 5}{3 x}+\frac{16 x / 5}{4 x}\right) \\ & =\frac{\mu_0 I 5}{48 \pi x}\left(\frac{3}{5}+\frac{4}{5}\right)=\frac{7 \mu_0 I}{48 \pi x} \therefore k=7 \end{aligned}$ ^

A steady current $I$ goes through a wire loop $P Q R$ having shape of a right angle triangle with $P Q=3 x,…

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